Two manned satellites approaching one another at a relative speed of 0.450 m/s i

Question

Two manned satellites approaching one another at a relative speed of 0.450 m/s intend to dock. The first has a mass of 2.50 103 kg, and the second a mass of 7.50 103 kg. Assume that the positive direction is directed from the second satellite towards the first satellite.

(a) Calculate the final velocity after docking, in the frame of reference in which the first satellite was originally at rest.


(b) What is the loss of kinetic energy in this inelastic collision?


(c) Repeat both parts, in the frame of reference in which the second satellite was originally at rest.
final velocity

m/s
loss of kinetic energy

Explanation / Answer

a)
7500kg *( 0.45m/s) + 0 = 2500kg * u + 7500kg * u
as the dock ,
3375 = 2500u + 7500u

=> u = 0.3375 m/s

initial kinetic energy = 0.5 *7500*0.45 2 = 759.37 J

final kinetic energy = 0.5 * ( 7500+ 2500) * ( 0.3375 ) 2=569.53 J

so change in energy = final - initial = -189.83 J

c) 2500kg *( 0.45m/s) + 0 = 2500kg * u + 7500kg * u
as the dock ,
1125 = 2500u + 7500u

=> u = 0.1125 m/s

initial kinetic energy = 0.5 *2500*0.45 2 = 253.125 J

final kinetic energy = 0.5 * ( 7500+ 2500) * ( 0.1125) 2=63.28 J

so change in energy = final - initial = -189.83 J

Because there are no
external forces, the velocity of the center of mass of the two-satellite system is unchanged by the collision.
The two velocities calculated above are the velocity of the center of mass in each of the two dierent individual
reference frames. The loss in KE is the same in both reference frames because the KE lost to internal forces
(heat, friction, etc.) is the same regardless of the coordinate system chosen.

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